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3 years ago

Why are hypothesis valuable even if it is not supported by data.

2 answers:

8 0

A hypothesis proposes a mechanism of change that can be tested. If the results indicate the hypothesized mechanism is not the one in operation it narrows the field and we know more. A falsified hypothesis tested by a well designed experiment still provides real evidence about the world.

It was once thought that maggots spontaneously arose from carrion. This was tested and falsified providing evidence this was not how new generations of flies appeared. This still left the question of how flies generated new life to be explored. One falsified hypothesis does not block further research into the question.

Even a failed hypothesis can increase scientific knowledge if done correctly. By knowing this hypothesis is not how something works, you save time for future scientists and can now use your effort in other investigations. That is the real value of hypothesis that are not supported by evidence.

disa [49]3 years ago

5 0

A hypothesis is an educated guess. Your data might not match your hypothesis because like i said, a hypothesis is a guess. Your data is multiple experiments to prove whether your guess was right or wrong. The hypothesis is valuable because because it gives you a reason to prove something right or wrong. Not sure how helpful that was but it's what i can think of...

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If sunlight begins to warm a frozen lake, the atoms in the lakes frozen water will begin to move faster. True of False?

Vinvika [58]

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4 years ago

A capacitor consists of two parallel plates, each with an area of 17.0 cm2 , separated by a distance of 0.150 cm . The material

Llana [10]

Answer:

a) $C = 40.138\,pF$ , b) $q = 16.056\,nC$ , c) $U = 3.212\,\mu J$

Explanation:

a) The capacitance of two parallel plates capacitor with dielectric is given by the following expression:

$C = K\cdot \epsilon_{o}\cdot \frac{A}{d}$

Where:

$K$ - Dielectric constant.

$\epsilon_{o}$ - Vaccum permitivity.

$A$ - Plate area.

$d$ - Distance between plates.

Hence, the capacitance of the system is:

$C = (4.00)\cdot (8.854\times 10^{-12}\,\frac{F}{m} )\cdot \left(\frac{17\times 10^{-4}\,m^{2}}{0.150\times 10^{-2}\,m}\right)$

$C = 4.014\cdot 10^{-11}\,F$

$C = 40.138\,pF$

b) The charge can be found by using the definition of capacitance:

$q = C\cdot V_{batt}$

$q = (4.014\times 10^{-11}\,F)\cdot (400\,V)$

$q = 1.606\times 10^{-8}\,C$

$q = 16.056\,nC$

c) The energy stored in the charged capacitor is:

$U=\frac{1}{2}\cdot Q\cdot V_{batt}$

$U=\frac{1}{2}\cdot (1.606\times 10^{-8}\,C)\cdot (400\,V)$

$U = 3.212\times 10^{-6}\,J$

$U = 3.212\,\mu J$

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